Week #15 - Word Problems & Differential Equations Section 8.1

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1 Week #15 - Word Problems & Differential Equations Section 8.1 From Calculus, Single Variable by Hughes-Hallett, Gleason, McCallum et. al. Copyright 25 by John Wiley & Sons, Inc. This material is used by permission of John Wiley & Sons, Inc. SUGGESTED PROBLEMS In Exercises 1-8, write a Riemann sum and then a definite integral representing the area of the region, using the strip shown. Evaluate the integral exactly Add vertical slices of height from x = to x = 5. Riemann Sum = x 5 Area = dx = x 5 = 15 Add up horizontal slices of length x = 1 y 2 from y = to y = 1. Riemann Sum = 1 y 2 y Exact Area = 1 1 y 2 dy. You can t evaluate this integral without a table, but we can figure out the area by the fact that it represents one quarter of a circle with radius 1, so Area = 1 4 π( 1) 2 = 5π 2. If you were asked to do such an integral, you would have been given Table Entry from the back of the text. 1

2 In Exercises 9-14, write a Riemann sum and then a definite integral representing the volume of the region, using the slice shown. Evaluate the integral exactly. (Regions are parts of cones, cylinders, spheres, and pyramids.) 9. We are adding up slices with cross-sectional area of πr 2 = π2 2 = 4π. The total volume would be represented by the sum of these slices: Riemann Sum = Area x = (4π) x. Volume = 9 4π dx = 4πx 9 = 6π This matches the standard volume formula for a cylinder, πr 2 h = 6π. 14. The construction of this volume integral is a little trickier than the others. First, we need to find the area of a horizontal slice y above the bottom of the pyramid. If at y =, the sides are all 2 long, and at y = 2, the sides are all long, we can use similar triangles to find that the sides at the general height y are l = 2 y long. The area of a cross-section at height y then is l 2 = (2 y) 2. Riemann Sum = Area y = (2 y) 2 y 2 Total volume = (2 y) 2 dy = 1 (2 2 y) = 1 [ (2 2) (2 ) ] = 8/ meters This agrees with the volume of a pyramid as the same as a cone, V = 1 ( base height ). 2

3 17. The integrals in Problems represent the area of either a triangle or part of a circle, and the variable of integration measures a distance. In each case, say which shape is represented, and give the radius of the circle or the base and height of the triangle. Make a sketch to support your answer showing the variable and all other relevant quantities. QUIZ PREPARATION PROBLEMS h 2 dh We are finding the area under the graph y = 15 h 2. This can be rewritten as y 2 = 15 h 2 or h 2 + y 2 = 15. This area therefore represents part of a circle of radius 15. Sketching a diagram, we see the area in question: y 15 h 2 15 h ( 5 1 h ) 7 dh In this question, we are finding the area under the function y = 1 h. This the formula 7 of a straight line, so we are most likely dealing with a triangle. We use the limits of integration to get the following sketch: y 5 5(1 h 7) 7 h

4 The represents the area of a triangle with base 7 and height Find, by slicing, the volume of a cone whose height is cm and whose base radius is 1 cm. Slice the cone as shown in Figure 8.4. If we take horizontal slices, let us define y as the height above the base. Then the radius decreases from r = 1 and y = to r = at y =. By similar triangle constructions, r = ( 1 y ) The volume of each slice would then be πr 2 y = π ( 1 y 2 ) y. Riemann Sum = ( π 1 y 2 y. Converting to an exact integral, ) Total Volume = ( π 1 y 2 dy ) = π 1 (1 y ) ( ) = π [ (1 1) (1 ) ] = π by substitution, w = y/ This agrees with the cone volume formula V = 1 ( base height ) = 1 (π(1)2 )() = π. 26. Figure 8.1 shows a solid with both rectangular and triangular cross sections. (a) Slice the solid parallel to the triangular faces. Sketch one slice and calculate its volume in terms of x, the distance of the slice from one end. Then write and evaluate an integral giving the volume of the solid. (b) Repeat part (a) for horizontal slices. Instead of x, use h, the distance of a slice from the top. Figure 8.1 (a) If we slice parallel to the triangles, then every cross-section is an identical triangle, like the one shown below. 4

5 2 The slices of width x then all have the same volume, 1 2 (2)() x. The total volume of the object is the sum of those slices, or x. In integral format, V = 4 dx = x 4 = 12 cubic centimeters. (b) We are taking horizontal slices in this case, of thickness h. The length of these slices is always 4 cm, but the width, w, will vary with the height of the slice, h. To determine the width, we look at the object end-on, facing the /2 triangle: w h Because the h/w triangle is similar to the /2 triangle, w/h = 2/ or w = 2h/. Thus the Riemann sum for the volume is ( ) 2h (rectangle area) h = (4 w) h = 4 h This converts to the integral form, with h going over the height of the object, from h = to h =. 5

6 Volume = ( ) 2h 4 dh h 2 = 8 2 = 4 [ 2 2] = 12 cubic centimeters This other way of slicing gives us a different form of volume integral, compared to part (a), but results in exactly the same final volume. 6

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